Solve (a-3m)^2-(3a+m)^2

$$( a - 3 m ) ^ { 2 } - ( 3 a + m ) ^ { 2 }$$

Evaluate

$$-4\left(2a-m\right)\left(2m+a\right)$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3m\right)^{2}.

$$a^{2}-6am+9m^{2}-\left(3a+m\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+m\right)^{2}.

$$a^{2}-6am+9m^{2}-\left(9a^{2}+6am+m^{2}\right)$$

To find the opposite of 9a^{2}+6am+m^{2}, find the opposite of each term.

$$a^{2}-6am+9m^{2}-9a^{2}-6am-m^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-6am+9m^{2}-6am-m^{2}$$

Combine -6am and -6am to get -12am.

$$-8a^{2}-12am+9m^{2}-m^{2}$$

Combine 9m^{2} and -m^{2} to get 8m^{2}.

$$-8a^{2}-12am+8m^{2}$$

Expand

$$8m^{2}-12am-8a^{2}$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-3m\right)^{2}.

$$a^{2}-6am+9m^{2}-\left(3a+m\right)^{2}$$

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+m\right)^{2}.

$$a^{2}-6am+9m^{2}-\left(9a^{2}+6am+m^{2}\right)$$

To find the opposite of 9a^{2}+6am+m^{2}, find the opposite of each term.

$$a^{2}-6am+9m^{2}-9a^{2}-6am-m^{2}$$

Combine a^{2} and -9a^{2} to get -8a^{2}.

$$-8a^{2}-6am+9m^{2}-6am-m^{2}$$

Combine -6am and -6am to get -12am.

$$-8a^{2}-12am+9m^{2}-m^{2}$$

Combine 9m^{2} and -m^{2} to get 8m^{2}.

$$-8a^{2}-12am+8m^{2}$$